Simplify and expand the following expression: $ \dfrac{2z - 9}{4z + 6}+\dfrac{z + 2}{z - 4} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(4z + 6)(z - 4)$ Multiply the first term by $\dfrac{z - 4}{z - 4}$ $ \begin{align*} \dfrac{2z - 9}{4z + 6} \times \dfrac{z - 4}{z - 4} & = \dfrac{(2z - 9)(z - 4)}{(4z + 6)(z - 4)} \\ & = \dfrac{2z^2 - 17z + 36}{(4z + 6)(z - 4)}\end{align*} $ Multiply the second term by $\dfrac{4z + 6}{4z + 6}$ $ \begin{align*} \dfrac{z + 2}{z - 4} \times \dfrac{4z + 6}{4z + 6} & = \dfrac{(z + 2)(4z + 6)}{(z - 4)(4z + 6)} \\ & = \dfrac{4z^2 + 14z + 12}{(z - 4)(4z + 6)}\end{align*} $ Now we have: $ = \dfrac{2z^2 - 17z + 36}{(4z + 6)(z - 4)} + \dfrac{4z^2 + 14z + 12}{(z - 4)(4z + 6)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{2z^2 - 17z + 36 + 4z^2 + 14z + 12}{(4z + 6)(z - 4)} $ $ = \dfrac{6z^2 - 3z + 48}{(4z + 6)(z - 4)}$ Expand the denominator: $ = \dfrac{6z^2 - 3z + 48}{4z^2 - 10z - 24}$